Expert: Paul Klarreich - 11/9/2008

Question
Differentiate.i) sin^3xcos3x + cos^3xsin3x.
ii) ln (sec x + tan x)

For part ii) can it just be the opposite of the integral, so it would be sec x?

2. Find the first and second derivatives (with respect to x) of the function

x= 4t^2 - 5t + 6 and y = t^3 - t^2 +t.


thank you very much for your help!

Answer
Questioner:   jenny
Category:  Advanced Math
Private:  No
 
Subject:  integration
Question:  Differentiate.
i) sin^3xcos3x + cos^3xsin3x.
ii) ln (sec x + tan x)

For part ii) can it just be the opposite of the integral, so it would be sec x?

2. Find the first and second derivatives (with respect to x) of the function

x= 4t^2 - 5t + 6 and y = t^3 - t^2 +t.


thank you very much for your help!
...........................
Hi, Jenny,

For i) sin^3xcos3x + cos^3xsin3x
I assume you mean:

sin^3(x) cos(3x) + cos^3(x)sin(3x)

for which you would want formulas for sin(3x) and cos(3x)

sin(3x) = 3 sin(x) - 4 sin^3(x),
cos(3x) = 4 cos^3(x) - 3 cos(x),

which you can look up or derive.

Now substitute and differentiate.

Is that what you meant?  If not, next time be more careful.

..............................

For ii) ln (sec x + tan x), use the chain rule, with

u = sec x + tan x

You get dy/du = 1/u

du/dx = sec x tan x + sec^2(x)

Now simplify.  Yes, you get  sec x, which is not the opposite of ln(sec x + tan x); the opposite of  ln (sec x + tan x) is
- ln (sec x + tan x)

I get the impression that you are careless about the vocabulary and about writing expressions.  That is a good way to fail math.

...............................

x= 4t^2 - 5t + 6 and y = t^3 - t^2 + t

I assume you want dy/dx and  d^2y/dx^2
       dy/dt   3t^2 - 2t + 1
dy/dx = ----- = -------------
       dx/dt     8t - 5
     
ddy     d  dy
---- = --- --
dxdx   dx  dx


d  dy     d          dy
-- --- = ---       -----
dx dx    dt(dx/dt)   dx

  d  dy
 --- --
 dt  dx
= --------
  dx/dt

         3t^2 - 2t + 1
  d/dt( --------------- )
             8t - 5
= --------------------------
       8t - 5

You can handle that.
[As you noted, you are getting d../dx, but you express your answers in terms of t.]